Calculating the nth super-root when n is greater than 2?

Question

Tetration (literally "4th operator iteration") is iterated exponentiation, much like how exponents are iterated multiplying. For example, $2$^^$3$ is the same as $2^{2^2}$, which is $2^4$ or 16. This applies to any two positive real numbers, and possibly even all real numbers.

There are two inverses of tetration, iterated logarithms "super-logarithms" or "tetralogarithms" and iterated roots "super-roots". For instance, the value of $\sqrt{256}_s$ is $4$ because $4$^^$2 = 4^4 = 256$.

It is very difficult to calculate with or approximate either of these inverse operations to tetration. Super-logarithms use some complicated recursive algorithms (see this wiki article), but super-roots are even harder; the example I showed just turned out to be easy-to-find because $256$ is a perfect super-square (that is, $4^4$). I could only manage to find a formula for super square roots on Wikipedia using the Lambert W function. I originally did not know what this Lambert W function was, but soon after I found out it's just the inverse function of $xe^x$. This expression stated that $ssrt(x) = ln(x)/W(ln(x))$.

However, I cannot find a formula or even approximation for higher super-roots at all. I tried using linear approximation along with a Euclidean distance version of linear approximation on $^3\sqrt{2}_s$ and both were very innacurate. I could eventually calculate $^3\sqrt{2}_s$ to 42 digits as $1.4766843373578699470892355853738898365517$ using this high-precision calculator and binary guess-and-check methods, but I still can't seem to find a generic formula or approximation for these super-roots which have orders higher than $2$, so I would like to know if anyone knows of one.

Answer 1

One thing we can do is render the third superroot as an iteration of second superroots.

Let $x^{x^x}=a$. Then

$(x^x)^{(x^x)}=a^x$

and we take two square superroots to get a fixed point iteration:

$\color{blue}{x=\sqrt{\sqrt{a^x}_s}_s}$

If we put $a>1$ and $x=1$ on the right side, we get

$x=\sqrt{\sqrt{a}_s}_s$

as our next iteration, and this will lie between $1$ and the true third superroot. This suggests that the scheme will be convergent.

Let's try this on the third superroot of $16$. Put $x=1$ on the right side of the fixed point relation above. Then we get

$x=\sqrt{\sqrt{16}_s}_s=1.769\text{ (4 digits)}$

Next time around we get to the same accuracy

$x=\sqrt{\sqrt{134.9}_s}_s=1.958\text{ (4 digits)}$

We seem to be converging to $2$ fairly rapidly.

For the third superroot of 2 the corresponding iterations are $1.000, 1.380, 1.459, 1.473, 1.476, \color{blue}{1.477},...$.

For smaller $a$ values the performance deteriorates, an outcome connected with square superroots being no longer single valued when the argument drops below 1. If $a=0.9$ and an initial value $x=1$ is tried, the iteration converges to about $0.890$ if we take square superroots $\ge 1/e$; but this does not work for $a=0.8$ due to one of the square superroots being undefined over the reals.

Answer 2

There is also an (individual) powerseries solution (Puisieux-series) for each $n$ separately. Unfortunately that series have a little radius of convergence (if nonzero at all), but might be summable using Euler-summation. I'll give an example for $n=3$. (More examples are in my small treatize on my webspace)

Let's define our basic function $$ v = f_3(u) = u \cdot \exp( u \cdot \exp(u))$$ This has a power series that Pari/GP would easily display to many, say $64$, coefficients. It is Lagrange-invertible, say $$ u = g_3(v) = g_{3,1} v + g_{3,2}/2! v^2 + g_{3,3}/3! v^3 + ... \qquad \qquad \\\ \qquad \qquad\qquad \qquad = v - 2/2! v^2 + 3/3! v^3 + 20/4! v^4 - 295/5! v^5 + 1554/6! v^6 + O(v^7) $$ Let's call the $n$'th superroot-function $r_n(y)$ such that $$ x = r_n(y) \to y = \;^n x $$ then with $n=3$ we have asymptotically (or with small range of convergence) $$ r_3(y) = \exp(g_3(\log(y))) $$

By visual impression using the first 64 coefficients it seems, as if $|y| \lt \exp(1/2)$ might give convergence. (Update: checked with $\small y=\exp(\exp(-1))\approx 1.444 $ and it seems to be computable without Euler-summation at all)
Using Euler-summation the radius of convergence seems to be extensible - but I do not have the exact estimate for the growth of the coefficients. Let's see the summation of $g_3(v)$ for $y=\exp(1/2)$ such that $v = \log(y) =1/2$

partial sums of series g_3(1/2)
index  direct par-   Eulersummati- x=exp(g_3(v)) 
       tial sums     on order 0.5
   0              0             0  1.000000000
   1   0.5000000000  0.2500000000  1.284025417
   2   0.2500000000  0.3125000000  1.366837941
   3   0.3125000000  0.3203125000  1.377558184
   4   0.3645833333  0.3196614583  1.376661628
   5   0.2877604167  0.3198649089  1.376941739
   6   0.3214843750  0.3205749512  1.377919773
   7   0.3465603299  0.3209987217  1.378503819
   8   0.2958108569  0.3211143463  1.378663217
   9   0.3245950850  0.3211192669  1.378670001
  10   0.3403287830  0.3211180191  1.378668280
  11   0.2981245348  0.3211251293  1.378678083
  12   0.3269177171  0.3211317745  1.378687245
  13   0.3373635102  0.3211344667  1.378690956
  14   0.2982079698  0.3211348582  1.378691496
 ...     ...             ...           ...
  55   0.1639008657  0.3211350702  1.378691789
  56   0.2430865079  0.3211350702  1.378691789
  57   0.5951181182  0.3211350702  1.378691789
  58  0.09819587958  0.3211350702  1.378691789
  59   0.2552736284  0.3211350702  1.378691789
  60   0.6589999027  0.3211350702  1.378691789
  61  0.01025715717  0.3211350702  1.378691789
  62   0.2806542398  0.3211350702  1.378691789
  63   0.7342007399  0.3211350702  1.378691789
  64  -0.1064953930  0.3211350702  1.378691789
  65   0.3257696112  0.3211350702  1.378691789
  66   0.8210482830  0.3211350702  1.378691789                

giving the result

\\ v = 1/2
\\ y = exp(v) = 1.648721271
\\ x = 1.378691789...  from the last entry in the above (exp() of Euler-summation)
x^x^x     \\ check result of exp(Eulersummation)
%368 = 1.648721271
x^x^x- y  \\ check error
%369 = -1.699715302 E-22

Using Euler-summation of higher order and still using $64$ coefficients I could approximate $x$ for $y=16$ thus $v=\log(16) \approx 2.772588722 $ to a couple of digits

 partial sums of series g_3(v)
index    direct par-   Eulersummati- x=exp(g_3(v)) 
         tial sums     on order 5.8
   0                0             0  1.000000000
   1      2.772588722  0.4138192123  1.512583646
   2     -4.914659500  0.5946280226  1.812356664
   3      5.742129363  0.6381957710  1.893062278
   4      54.98695039  0.6360439550  1.888993136
   5     -347.7931741  0.6382424852  1.893150713
   6      632.6698779  0.6531183317  1.921523444
   7      4675.314968  0.6695911376  1.953438470
   8     -40693.25817  0.6786063818  1.971128816
   9      101996.7689  0.6804959536  1.974856927
  10      534495.9511  0.6805394727  1.974942873
  11     -5898689.882  0.6823339559  1.978490056
  12      18438758.44  0.6856034795  1.984969362
  13      67398852.83  0.6883322388  1.990393263
  14     -950277060.1  0.6894759540  1.992671008
  ...      ...            ...           ...
  55  -2.664574273E40  0.6931453682  1.999996375
  56   9.487882296E39  0.6931457660  1.999997171
  57   9.002509460E41  0.6931461912  1.999998021
  58  -6.072183170E42  0.6931464204  1.999998480
  59   6.149361593E42  0.6931464569  1.999998553
  60   1.803356223E44  0.6931464662  1.999998571
  61  -1.371747030E45  0.6931465681  1.999998775
  62   2.215487571E45  0.6931467349  1.999999109
  63   3.558078850E46  0.6931468656  1.999999370
  64  -3.073667769E47  0.6931469109  1.999999461
  65   6.704426035E47  0.6931469096  1.999999458
  66   6.882982587E48  0.6931469259  1.999999491

We see (and conclude for the general trend), that direct evaluation of the series $g_3(v)$ leads to unbounded partial sums, but which might be handled successfully by Eulersummation of appropriate order. For negative $v$ this might look different because Euler-summation can only be successful if the summands-to-be-partial-summed are (at least roughly) alternating in signs.

---

The same procedure can be done for higher $n$, getting simply different power series, and for $n \to \infty$ we get a well known powerseries with simple set of coefficients, see my small treatize which I have linked to.

---

update 2 Just for fun I tried $y=256$, $v \approx 5.545177444$. Using Eulersummation of order $11.37$ I got the following, with now $128$ terms of the power series needed:

partial sums of series g_3(v)
index      direct par-   Eulersummati- x=exp(g_3(v)) 
           tial sums     on order 11.37
    0                 0             0  1.000000000
    1       5.545177444  0.4482762688  1.565611165
    2      -25.20381545  0.6593619374  1.933558210
    3       60.05049546  0.7137176680  2.041567036
    4       847.9676320  0.7103553682  2.034714203
    5      -12040.99635  0.7136655574  2.041460652
    6       50708.63898  0.7376632728  2.091043603
    7       568167.2105  0.7661483392  2.151463567
    8      -11046187.51  0.7827563692  2.187493503
    9       62011106.34  0.7863107520  2.195282527
   10       504890269.0  0.7863264545  2.195316998
   11   -1.267027432E10  0.7906504964  2.204830194
   12    8.701591401E10  0.7990721357  2.223476886
   13    4.880970073E11  0.8065436624  2.240151869
   14   -1.618550515E13  0.8098085843  2.247477743
  ...        ...              ...          ...
  115   6.351017290E120  0.8292221643  2.291535609
  116   1.111485492E123  0.8292222056  2.291535703
  117  -1.527759513E124  0.8292223349  2.291536000
  118   3.224076332E124  0.8292224524  2.291536269
  119   1.850204173E126  0.8292224818  2.291536336
  120  -2.865500371E127  0.8292224486  2.291536260
  121   9.519211594E127  0.8292224379  2.291536236
  122   3.024663081E129  0.8292224940  2.291536364
  123  -5.316849919E130  0.8292225794  2.291536560
  124   2.371536870E131  0.8292226279  2.291536671
  125   4.834907255E132  0.8292226204  2.291536654
  126  -9.764577483E133  0.8292225984  2.291536603
  127   5.423935796E134  0.8292226105  2.291536631
  128   7.506467361E135  0.8292226592  2.291536743
  129  -1.775490700E137  0.8292227055  2.291536849
  130   1.177132710E138  0.8292227171  2.291536875

Meaning the result

\\ y = 256
\\ v = log(y) ~ 5.545
\\ x = 2.291536875...  from the last entry in the above (exp() of Euler-summation)
x^x^x     \\ check result of exp(Eulersummation)
%465 = 255.9990213

Using $256$ series coefficients and increased internal precision I got the result correct to even the eleventh digit.
So this as an initial solution and then Newton-interpolation (perhaps somehow following the idea of Oscar Lanzi) one might arrive at arbitrary accuracy - again: I did not check the lower bound for $y$...

Answer 3

Just for sake of simplicity, it is possible to compute $\sqrt[n]a_s$ very easily when $a>1$ or $n$ is odd and $a>0$ using bisection, which doesn't involve any hard calculations (so this is something doable with a calculator that only has exponentiation, and perhaps some paper).

As long as you have $x\le\sqrt[n]a_s\le y$, we can iteratively consider $[(x+y)/2]\widehat~\widehat~n$, and replace either $x$ or $y$ with $(x+y)/2$.

To find initial values to work with, note that $a\ge\sqrt[n]a_s$. From there, repeatedly divide by 2 until you find a value where $x\le\sqrt[n]a_s$, and set $y$ to be $2x$. Then apply the above.

Here is a simple program that implements this. It takes 53 steps to compute the result

$$\sqrt[3]2_s\approx1.4766843373578697$$

It is also notable that the above program may fail on larger inputs, since computing things like 256^^3 may not be reasonably doable. For a human, however, if the radicand is small enough, it is easy to find initial values manually.

Since some programming languages give $a^b=\infty$ if $a$ and $b$ are too large, and $\infty$ is considered greater than anything, there is no need to worry about larger inputs. This allows for the computation of things such as $\sqrt[7]{256}_s$ in only 60 steps. See here.

---

Improvement:

One can employ a faster root-finding algorithm that works well against functions with large second derivatives to get faster practical use. The idea is similar to the above, where we define a new function

$$f(x)=x\widehat~\widehat~n-a$$

which we want to find the root of. We use the initial boundaries

$$(x_l,x_u)=(\min(1,a),\max(1,a))$$

or $x_u=2$ if $f(2)>0$, and then iteratively define the bisection and regula falsi midpoints:

$$x_\mathrm B:=\frac{x_u+x_l}2$$

$$x_\mathrm{RF}:=\frac{x_uf(x_l)-x_lf(x_u)}{f(x_l)-f(x_u)}=\frac{\frac{x_u}{f(x_u)}-\frac{x_l}{f(x_l)}}{\frac1{f(x_u)}-\frac1{f(x_l)}}\stackrel{f(x_u)\to\infty}\longrightarrow x_l$$

where $x_\mathrm{RF}:=x_l$ is used if $f(x_u)$ is infinite in double precision. We then define our new endpoint as

$$x_r:=\begin{cases}\frac12(x_\mathrm B+x_\mathrm{RF}),&\text{steps}\bmod3\ne0\text{ or }x_u-x_l>0.5\text{ or }|f(x_\mathrm{RF})|>0.5\\x_\mathrm{RF},&\text{else}\end{cases}$$

$$x_l:=x_r\text{ if }f(x_r)<0$$

$$x_u:=x_r\text{ if }f(x_r)>0$$

which converges roughly twice as fast as bisection at worst and slightly faster than linear convergence asymptotically. See here for an implementation of the above, which computes $\sqrt[3]2_s$ and $\sqrt[7]{256}_s$ in 9 and 21 iterations respectively.

Answer 4

Just a long comment, no answer

@SimplyBeautifulMath's simple answer triggered me to have another look at the searching/solver ansatz.
It is not so elegant as SBM's procedere, but fast enough and also easy. Here is the Pari/GP-code

itet(b,h=1)=my(x=1);for(k=1,h,x=b^x);x  \\ integer tetration to height h

{ sroot(m,maxorder=1)=my(logm,vsroots,ub);
    logm=log(m) ;              \\ make the log a constant
    vsroots=vectorv(maxorder); \\ column-vector of superroots of order 1 to (maxorder)
    vsroots[1]=m;              \\ define sr of order 1 to be m 
    vsroots[2]= exp(LambertW(logm));  \\ use LambertW for direct estimate
                               \\            for sr of order 2
    for(o=3,maxorder,          \\ for all requested higher orders...
            ub=vsroots[o-1];   \\ use previous sr as upper bound for current sr
            vsroots[o]=solve(sr=1,ub, log(sr) * itet(sr,o-1)-logm) \\ solve 
                               \\ for identity of log (sr^^o) = log(m) 
        );
    return(vsroots);}
\\ do the computation to get a table of 20 x 10 entries.
tmp= Mat(vector(10,e,sroot(10^e,20)~))

The idea is here to use a "good" estimate for the upper-bound for $sr$ of order $o$ to avoid numerical overflow in the search-routine - just take the $sr$ of smaller order. Using $\log(sr \uparrow \uparrow o) = \log(m)$ reduces again the problem of numerical overflow.

This made the following table for 1 to 20'th superroot of powers of 10.

 order       m --> 
 ----+---------------------------------------------------------------------------------- ------------------
   1         10        100       1000      10000     100000    1000000   10000000  100000000  1000000000  10000000000
   2  2.5061841  3.5972850  4.5555357  5.4385827  6.2709196  7.0657967  7.8313895  8.5731845   9.2950869    10.000000
   3  1.9235840  2.2127958  2.3849098  2.5072341  2.6019716  2.6791912  2.7443082  2.8005682   2.8500690    2.8942439
   4  1.7343125  1.8662152  1.9319788  1.9743344  2.0050183  2.0288114  2.0480991  2.0642312   2.0780410    2.0900767
   5  1.6440791  1.7188182  1.7522674  1.7726883  1.7869853  1.7978034  1.8064095  1.8134996   1.8194934    1.8246618
   6  1.5924155  1.6399925  1.6597509  1.6714095  1.6794032  1.6853641  1.6900542  1.6938845   1.6970995    1.6998551
   7  1.5594683  1.5920497  1.6048495  1.6122253  1.6172116  1.6208940  1.6237705  1.6261064   1.6280579    1.6297241
   8  1.5369070  1.5603758  1.5692069  1.5742077  1.5775542  1.5800085  1.5819159  1.5834587   1.5847435    1.5858375
   9  1.5206559  1.5382042  1.5445837  1.5481481  1.5505152  1.5522422  1.5535793  1.5546577   1.5555535    1.5563148
  10  1.5084979  1.5220043  1.5267778  1.5294168  1.5311588  1.5324247  1.5334019  1.5341882   1.5348403    1.5353935
  11  1.4991297  1.5097690  1.5134418  1.5154549  1.5167774  1.5177354  1.5184732  1.5190657   1.5195564    1.5199723
  12  1.4917381  1.5002805  1.5031713  1.5047446  1.5057741  1.5065179  1.5070896  1.5075482   1.5079274    1.5082485
  13  1.4857915  1.4927612  1.4950799  1.4963344  1.4971526  1.4977424  1.4981951  1.4985577   1.4988573    1.4991108
  14  1.4809287  1.4866938  1.4885837  1.4896010  1.4902626  1.4907387  1.4911036  1.4913956   1.4916367    1.4918405
  15  1.4768968  1.4817223  1.4832840  1.4841209  1.4846640  1.4850541  1.4853528  1.4855916   1.4857886    1.4859551
  16  1.4735134  1.4775948  1.4789007  1.4795980  1.4800494  1.4803733  1.4806210  1.4808190   1.4809821    1.4811200
  17  1.4706445  1.4741283  1.4752319  1.4758191  1.4761987  1.4764707  1.4766785  1.4768445   1.4769812    1.4770967
  18  1.4681892  1.4711875  1.4721288  1.4726282  1.4729505  1.4731812  1.4733573  1.4734979   1.4736137    1.4737114
  19  1.4660708  1.4686702  1.4694798  1.4699082  1.4701842  1.4703816  1.4705323  1.4706524   1.4707513    1.4708348
  20  1.4642296  1.4664982  1.4671997  1.4675700  1.4678083  1.4679786  1.4681085  1.4682120   1.4682972    1.4683691

As already said, I find of course the procedere as shown by SBM much more elegant and also more interesting. This comment is just to have a table of values visible.

Answer 5

$\def\srt{\operatorname{srt}}$ Although there are integral solutions, the super root $\sqrt[n]x_s,\srt_n(x)$ has an expansion using Lagrange reversion:

$$\sqrt[k]x_s=\srt_k(x)= 1+\sum_{n=1}^\infty\frac{\ln^n(x)}{n!}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0$$

Applying general Leibniz rule and $e^z$ Maclaurin expansion each $k-2$ times if $2<k\in\Bbb N$:

$$ \frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^\infty\sum_{n_2=0}^{n-1}\dots\sum_{n_{2k-5}=0}^\infty\sum_{n_{2k-4}=0}^{n-1-\sum_\limits{j=1}^{k-3}n_{2j}}\left.\frac{d^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}{dt^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}e^{(n_{2k-5}+1)t}\right|_0\prod_{j=1}^{k-2}\frac{n_{2j-3}^{n_{2j-1}}}{n_{2j-1}!}\left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_{2m}}\\n_{2j}\end{matrix}\right)\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0$$

where $n_{-1}=-n$. Next, use Kronecker delta and factorial power $m^{(n)}$ in $\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0=\delta_{n_{2j},n_{2j-1}}n_{2j-1}^{(n_{2j})}$. As hinted by Quantile Mechanics $(96)$ to $(97)$, remove the odd indexed sums substituting each $n_{2j}=n_{2j-1}$. We reindex $n_{2j}\to n_j$ and simplify:

$$\begin{align}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^{n-1}\dots \sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j} (n_{k-2}+1)^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-2}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)= \sum_{n_1=0}^{n-1}\dots \sum_{n_{k-1}=0}^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-1}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)\end{align}$$

where $n_0=n$. These links support the last equality. Therefore:

$$\bbox[1px, border: 2px solid red]{\sqrt[k]z_s=\srt_k(z)=1+\sum_{n=1}^\infty \sum_{n_1=0}^{n-1}\dots\sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j}\frac{(-1)^{n_1}\ln^n(z)}{\Gamma \left({n-\sum_\limits{j=1}^{k-2}n_m}\right)(n_{k-2}+1)^{1-n+\sum\limits_{j=1}^{k-2}n_j} n}\prod_{j=1}^{k-2}\frac {n_{j-1}^{n_j}}{n_j!}}$$

where all can be infinite sums or all, except the $n$ sum, can have an upper bound of $n$. Assume improper sums are empty. We find that:

$$\srt_1(z)=z$$ $$\srt_2(z)=1+\ln(z)+\sum_{n=2}^\infty\frac{\ln^n(z)}{n!}(1-n)^{n-1}$$ $$\srt_3(z)=1-\sum_{n=1}^\infty\sum_{k=1}^{n-2}\frac{(-1)^kk^{n-k+1}n^{k-2}\ln^n(z)}{(n-k)!k!}$$

shown here

$$\srt_4(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=1}^{n-k-1}\frac{(-1)^k n^{k-1}k^{m-1}\ln^n(z)}{(n-k-m)!k!m!m^{m+k-n-1}}$$

Shown here

$$\srt_5(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{j=1}^{n-k-m-1}\frac{(-1)^k k^m n^{k-1} m^{j-1}\ln^n(z)}{(n-j-k-m)!m!k!j!j^{j+m+k-n-1}}$$ Shown here

$$\srt_6(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-1-k}\sum_{j=0}^{n-1-k-m}\sum_{p=1}^{n-k-m-j}\frac{(-1)^k n^{k-1}k^m m^j j^{p-1}\ln^n(z)}{(n-1-k-m-j-p)!k!m!j!p!p^{p+j+m+k-n-1}}$$ shown here $$\vdots$$ $$\srt_\infty(z)=\sqrt[z]z$$

Compare both links for verification. The region of convergence is near $|z|=1$. Help is wanted with the region of convergence for the multiple series expansion and other representations of $\srt_k(z)$.